Question

question 9 (6 points)
the equilibrium expression for kp for the reaction below is
2o3 (g) ⇌ 3o2 (g)

Explanation:

Step1: Recall Kp expression rule

For a reaction \( aA(g)
ightleftharpoons bB(g) \), \( K_p=\frac{(P_B)^b}{(P_A)^a} \), where \( P_A, P_B \) are partial pressures.

Step2: Apply to given reaction

Reaction: \( 2O_3(g)
ightleftharpoons 3O_2(g) \) (assuming typo, correct formula).
Here, \( a = 2 \) (for \( O_3 \)), \( b = 3 \) (for \( O_2 \)).
Thus, \( K_p=\frac{(P_{O_2})^3}{(P_{O_3})^2} \).

Answer:

\(\boldsymbol{\frac{(P_{O_2})^3}{(P_{O_3})^2}}\) (matching option with this form, e.g., if option a is \(\frac{3P_{O_2}}{2P_{O_3}}\) no, correct is \(\frac{(P_{O_2})^3}{(P_{O_3})^2}\); assuming options have exponents, the correct \( K_p \) is ratio of product partial pressure (raised to its coefficient) over reactant partial pressure (raised to its coefficient)).